Algebra II

Chapter 10 Section 5Hyperbola

Hyperbola

The set of all points that have the difference of the distance to 2 focus points is a given constant

Parts of a hyperbola

Transverse Axis

Foci

Asymptotes

Vertices

Standard Form of an Hyperbola

If a & b > 0 the Transverse axis is Horizontal

If a & b < 0 the Transverse axis is Vertical

12

2

2

2

b

y

a

x

Put into standard Form

3649 22 yx

Put into standard Form

324936 22 yx

In standard form• The +-value for a (square root of the

denominator of x2) are the vertex’s x if the Transversal Axis is horizontal, values (-a,0) and (a,0)

• The +-value for b (square root of the denominator of y2) are the vertex’s y if the Transversal Axis is Vertical, values (0,-b) and (0,b)

To find the Asymptote

xa

by

xa

by

Find the Asymptotes and Vertices

12516

22

yx

1949

22

xy

12516

22

yx

xa

by

Horizontal Transverse Axis

So Vertices are at +- (a,0)

a = 4 b = 5

(4,0) and (-4,0)

Asymptotes are at

xy4

5 xy

4

5

1949

22

xy

Given a vertex at (-3, 0) and an asymptote of y = 4x

Find the equation of hyperbola

Foci

• Foci are always on the transverse axis

• Foci are equidistant from the minor axis

• The distances to a focus from the origin is equal to the distance along the asymptote, with a coordinate = to the vertex’s non zero coordinate

Find the focus of an Hyperbola with its center on the origin

• If c is the distance from the center to the focus

• Find c using c2 = a2 + b2

• if your transverse axis is Horizontal your foci are at (+-c ,0)

• if your transverse axis is Vertical your foci are at (0 ,+-c)

a & b are + so we have a horizontal Transverse axis (y = 0)

c2 = a2 + b2

c2 = 16 + 9c2 = 25c = 5

So the coordinates of the foci are at (5,0) and (-5,0)

Find the foci of 1

916

22

yx

125144

22

xy

Find the Foci

Find the equation given a vertex and a focus

• Find the other vertex using c2 = a2 + b2

• Substitute in the 2 values you know and solve for the 3rd

– Remember the foci will lie along the Transverse axis

– c is the distance of the focus to the origin

• Once you have the 2 vertices, take the square roots to get a and b & substitute them in

Given a focus of (10,0) and a vertex at (-6,0) find the equation of the

Hyperbola

The transverse axis is Horizontal since the vertex and focus lie on the x axis

Since the vertex is at (0,-6) a2 = 36Since the focus is at (0,10) c2 = 100

Substitute a2 and c2 into a2 + b2 = c2 and solve for b36 + b2 = 100 b2 = 64 b= 8

Substitute a2 & b2 into 12

2

2

2

b

y

a

x1

6436

22

yxTo

get

Given a focus of (13,0) and a vertex at (5,0) find the equation of the ellipse

Given a focus of (0,4) and a vertex at (0,5) find the equation of the ellipse

Do Now!

• Page 566 Problems 2 – 18 even, find the vertex, foci and

asymptotes as well as sketch a graph

23 – 26 all30, 31, 36, 40, 42, 49-51